Đặt \(P=\frac{9x^2-12x+4}{x^2-2x+2}\)
\(=\frac{10\left(x^2-2x+2\right)-x^2+8x-16}{x^2-2x+2}\)
\(=\frac{10\left(x^2-2x+2\right)}{x^2-2x+2}-\frac{x^2-8x+16}{x^2-2x+2}\)
\(=10-\frac{\left(x-4\right)^2}{x^2-2x+2}\)
Ta thấy : \(\frac{\left(x-4\right)^2}{x^2-2x+2}\ge0\left(x^2-2x+2=\left(x-1\right)^2+1,\left(x-4\right)^2\ge0\right)\)
\(\Rightarrow10-\frac{\left(x-4\right)^2}{x^2-2x+2}\le10\)
hay \(P\le10\)
Dấu "=" xảy ra \(\Leftrightarrow x=4\)
Vậy : GTLN của \(P=10\Leftrightarrow x=4\)